Odpowiedź:
[tex]R\'ownanie\ \ okregu\\\\(x-a)^2+(y-b)^2=r^2\ \ \ \ \ \ S=(a,b)\ \ -wsp\'olrzedne\ \ \'srodka\\\\\\a)\ \ S=(-1,3)\ \ \ \ \ \ r=3\\\\(x-(-1))^2+(y-3)^2=3^2\\\\(x+1)^2+(y-3)^2=9\\\\\\b)\ \ S=(0,0)\ \ \ \ \ \ r=11\\\\(x-0)^2+(y-0)^2=11^2\\\\x^2+y^2=121\\\\\\c)\ \ S=(-2,-5)\ \ \ \ \ \ r=1\\\\(x-(-2))^2+(y-(-5))^2=1^2\\\\(x+2)^2+(y+5)^2=1\\\\\\d)\ \ S=(0,5)\ \ \ \ \ \ r=3\sqrt{2}\\\\(x-0)^2+(y-5)^2=(3\sqrt{2})^2\\\\x^2+(y-5)^2=9\cdot2\\\\x^2+(y-5)^2=18[/tex]
[tex]e)\ \ S=(6,-4)\ \ \ \ \ \ r=5\\\\(x-6)^2+(y-(-4))^2=5^2\\\\(x-6)^2+(y+4)^2=25\\\\\\f)\ \ S=(-\frac{1}{3},0)\ \ \ \ \ \ r=\sqrt{3}\\\\(x-(-\frac{1}{3}))^2+(y-0)^2=(\sqrt{3})^2\\\\(x+\frac{1}{3})^2+y^2=3[/tex]
